Sarah Alanazi
Jacan Chaplais
April, 2021
In 1927, Dirac proposed that quantum mechanics may be elevated to a relativistic theory by quantizing a spinor field [1].
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During this calculation, we shall make use of two different inertial reference frames to obtain our final result. Mandelstam variables are constructed from the incoming and outgoing momenta of the interaction. \[ \begin{array}{lllllllll} s &=& (p+k)^{2} &=& p^{2}+k^{2}+2 p \cdot k &=& m^{2}+2 p \cdot k &=& m^{2}+2 p^{\prime} \cdot k^{\prime} \\ t &=& \left(p^{\prime}-p\right)^{2} &=& p^{\prime 2}+p^{2}-2 p \cdot p^{\prime} &=& 2 m^{2}-2 p \cdot p^{\prime} &=& -2 k \cdot k^{\prime} \\ u &=& \left(k^{\prime}-p\right)^{2} &=& k^{\prime 2}+p^{2}-2 k^{\prime} \cdot p &=& m^{2}-2 k^{\prime} \cdot p &=& m^{2}-2 k \cdot p^{\prime} \end{array} \] Casting transition amplitudes in terms of these manifestly Lorentz invariant quantities allows us to jump between frames with ease.
Inetial frame in which sum of spatial momenta is zero.
Inertial frame in which the electron is at rest. This will also be the assumed rest frame for our particle detectors, hence lab frame.
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\[ \begin{aligned} \mathrm{d} \sigma=& \frac{1}{2 E_{\mathcal{A}} 2 E_{\mathcal{B}}\left|v_{\mathcal{A}}-v_{\mathcal{B}}\right|}\left(\prod_{f} \frac{\mathrm{d}^{3} p_{f}}{(2 \pi)^{3}} \frac{1}{2 E_{f}}\right) \\ &\left|\mathcal{M}\left(p_{\mathcal{A}}, p_{\mathcal{B}} \rightarrow\left\{p_{f}\right\}\right)\right|^{2}(2 \pi)^{4} \delta^{(4)}\left(p_{\mathcal{A}}+p_{\mathcal{B}}-\sum p_{f}\right) \end{aligned} \]
\[ \int \mathrm{d} \Pi_{2}=\int \frac{\mathrm{d}^{3} k^{\prime}} {(2 \pi)^{3} 2 E_{k^{\prime}}} \frac{\mathrm{d}^{3} p^{\prime}}{(2 \pi)^{3} 2 E_{p^{\prime}}}(2 \pi)^{4} \delta^{4}\left(p+k-k^{\prime}-p^{\prime}\right) \]
By applying the Feynman rules to these diagrams and grouping terms, we obtain \[ i \mathcal{M}=i e^{2} \epsilon_{\mu \lambda^\prime}^{\ast} \left(k^{\prime}\right) \epsilon_{\nu \lambda}\left(k\right) \bar{u}^{s^{\prime}}\left(p^{\prime}\right) \left( \dfrac{% numerator \gamma^{\mu}(\not{p}+k+m) \gamma^{\nu}} {(p+k)^{2}-m^{2}}% denominator + \dfrac{% numerator \gamma^{\nu} \left(\not{p}-k^{\prime}+m\right) \gamma^{\mu}} {\left(p-k^{\prime}\right)^{2}-m^{2}}% denominator \right) u^{s}(p) \]
This unwieldy expression can be simplified a little by expanding the binomials in the denominator, and observing for the numerator \[ \begin{aligned} (\not{p}+m) \gamma^{\nu} u^{s}(p) &=\left(2 p^{\nu}-\gamma^{\nu} \not{p}+\gamma^{\nu} m\right) u^{s}(p) \\ &=2 p^{\nu} u^{s}(p)-\gamma^{\nu}\underbrace{(\not{p}-m) u^{s}(p)}_{ \text{Dirac equation} \implies 0 } \\ &=2_{} p^{\nu} u^{s}(p) \end{aligned} \]
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[1] Dirac PAM. Quantum theory of emission and absorption of radiation. Proc Roy Soc Lond A 1927;114:243. https://doi.org/10.1098/rspa.1927.0039.